= Solution
In the convention of <4/a>, the <fundamental weights> are
$$
\omega_1=e_1,\qquad\omega_2=e_1+e_2,\qquad
\omega_3=\tfrac12(e_1+e_2+e_3).
$$
For a <weight> $w=(w_1,w_2,w_3)$ in this orthonormal coordinate system, its <Dynkin labels> are
$$
\boxed{a=(a_1,a_2,a_3)=(w_1-w_2,\ w_2-w_3,\ 2w_3).}
$$
These labels, rather than the orthonormal coordinates, are used below and in the projection in <4/c>.
The <highest weight> $(1,0,0)$ is $e_1$. Successive subtraction of the <simple roots> gives the complete string
$$
e_1\xrightarrow{-\alpha_1}e_2\xrightarrow{-\alpha_2}e_3
\xrightarrow{-\alpha_3}0\xrightarrow{-\alpha_3}-e_3
\xrightarrow{-\alpha_2}-e_2\xrightarrow{-\alpha_1}-e_1.
$$
The two consecutive short-root steps are allowed because $e_3$ has coroot pairing two with $\alpha_3$. The <weight system> of this seven-dimensional vector <representation> is therefore
$$
\boxed{\begin{array}{c|r}
w& (a_1,a_2,a_3)\\\hline
e_1&(1,0,0)\\
e_2&(-1,1,0)\\
e_3&(0,-1,2)\\
0&(0,0,0)\\
-e_3&(0,1,-2)\\
-e_2&(1,-1,0)\\
-e_1&(-1,0,0)
\end{array}.}
$$
All seven <weight multiplicities> are one, including the zero <weight>.
The <highest weight> $(0,0,1)$ is $\omega_3$. The <Weyl group> of the <B3 root system> acts by signed coordinate permutations. Its orbit is the eight cube vertices $w=(s_1e_1+s_2e_2+s_3e_3)/2$, with independent $s_i=\pm1$. Starting from $(+,+,+)$, lowering by $\alpha_3$, then $\alpha_2$, and then either $\alpha_1$ or $\alpha_3$ generates the middle branches; further lowering reaches all remaining sign choices. Thus the eight-dimensional <spin representation> has
$$
\boxed{\begin{array}{c|r}
(s_1,s_2,s_3)&(a_1,a_2,a_3)\\\hline
(+,+,+)&(0,0,1)\\
(+,+,-)&(0,1,-1)\\
(+,-,+)&(1,-1,1)\\
(+,-,-)&(1,0,-1)\\
(-,+,+)&(-1,0,1)\\
(-,+,-)&(-1,1,-1)\\
(-,-,+)&(0,-1,1)\\
(-,-,-)&(0,0,-1)
\end{array}.}
$$
The stipulated unit <weight multiplicities> give dimensions seven and eight by counting these lists.
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